Question: $\sum\limits_{k=0}^{{19}}{{{-4(-2)^{k}}}} \approx$ Choose 1 answer: Choose 1 answer: (Choice A) A $ -699{,}052 $ (Choice B) B $ 1{,}398{,}100 $ (Choice C) C $2{,}097{,}152$ (Choice D) D $ 4{,}194{,}308 $
Solution: What is the question asking for? The question is asking for the sum of the values of $-4(-2)^k$ from $k = 0$ to $k = 19$ : $-4(-2)^0 -4(-2)^1 -... -4(-2)^{19} $ The series is geometric because the formula $-4(-2)^k$ is an exponential function of $k$. Formula for geometric series The sum $S_n$ of a finite geometric series is $S_n = \dfrac{a_1(1-r^n)}{1-r}$ where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. What do we need to use the formula? The number of terms $n$ is ${20}$ because there are ${20}$ numbers from $0$ to $19$. The first term $a_1$ is ${-4}$ because $-4(-2)^0 = {-4}$. The common ratio $r$ is ${-2}$ because it is the base of the exponent in $-4({-2})^k$. Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac{a_1(1-r^n)}{1-r} \\\\ S_{{20}}&=\dfrac{{-4}(1-\left({-2}\right)^{{20}})}{1-\left({-2}\right)} \\\\ S_{{20}}&=\dfrac{-4}{3}\left(1-\left({-2}\right)^{{20}}\right) \\\\\\ S_{{{20}}} &\approx 1{,}398{,}100 \end{aligned}$ The answer $ 1{,}398{,}100 $